Determine where the function f has a derivative, as a function of a complex variable:
f(x +iy) = 1/(x+i3y)
I know the cauchy-riemann is not satisfied, so does that simply mean the function is not differentiable anywhere?
Determine where the function f has a derivative, as a function of a complex variable:
f(x +iy) = 1/(x+i3y)
I know the cauchy-riemann is not satisfied, so does that simply mean the function is not differentiable anywhere?
Alright, so doing the partials and equating I get
x = 3y and
xy = 3xy
So, subbing first equation into the second we get,
3y^2=9y^2, so y=0. If y=0 then x=0. So differentiable only at 0+0i.
But, because 0 + i0 is undefined in the original function, would this be differentiable nowhere?
Am I approaching this right?
Great..
so just to clarify, in general there are cases where cauchy-riemann are not satisfied, however derivatives do exisit at some points? (In Other Words, if Cauchy-Riemann is not satisfied, it does not neccessarily mean the function isnt differentiable everywhere?)
correct (if you're saying what i think you're saying), and also, the equations can be satisfied, but the function is still not differentiable. The Cauchy-Riemann is a necessary but not sufficient condition for differentiability (i don't like how you phrased what you said though, the equations must be satisfied for differentiability, but the formulas don't have to look the same, that's what i'm saying, you just have to check what points work, if any)
It is not clear to me from the discussion that the function was ever put in proper u+iv form.
The function must be rewritten as $\displaystyle \frac{1}{{x + 3yi}} = \frac{x}{{x^2 + 9y^2 }} + i\frac{{ - 3y}}{{x^2 + 9y^2 }}$.
So $\displaystyle u(x,y) = \frac{x}{{x^2 + 9y^2 }}\;\& \;v(x,y) = \frac{{ - 3y}}{{x^2 + 9y^2 }}$.
That may have been done, but it is clear. Was it done?